# 684. Redundant Connection 冗余连接

@TOC

## # 题目描述

In this problem, a tree is an undirected graph that is connected and has no cycles.

The given input is a graph that started as a tree with N nodes (with distinct values 1, 2, ..., N), with one additional edge added. The added edge has two different vertices chosen from 1 to N, and was not an edge that already existed.

The resulting graph is given as a 2D-array of `edges`. Each element of `edges` is a pair `[u, v]` with u < v, that represents an undirected `edge` connecting nodes u and v.

Return an `edge` that can be removed so that the resulting graph is a tree of N nodes. If there are multiple answers, return the answer that occurs last in the given 2D-array. The answer edge `[u, v]` should be in the same format, with u < v.

``````Example 1:

Input: [[1,2], [1,3], [2,3]]
Output: [2,3]
Explanation: The given undirected graph will be like this:
1
/ \
2 - 3

Example 2:

Input: [[1,2], [2,3], [3,4], [1,4], [1,5]]
Output: [1,4]
Explanation: The given undirected graph will be like this:
5 - 1 - 2
|   |
4 - 3
``````

Note:

1. The size of the input 2D-array will be between 3 and 1000.
2. Every integer represented in the 2D-array will be between 1 and N, where N is the size of the input array.

## # 解题方法

### # 并查集

python代码如下：

``````class Solution:
def findRedundantConnection(self, edges):
"""
:type edges: List[List[int]]
:rtype: List[int]
"""
tree = [-1] * (len(edges) + 1)
for edge in edges:
a = self.findRoot(edge[0], tree)
b = self.findRoot(edge[1], tree)
if a != b:
tree[a] = b
else:
return edge

def findRoot(self, x, tree):
if tree[x] == -1: return x
else:
root = self.findRoot(tree[x], tree)
tree[x] = root
return root
``````

C++代码如下：

``````class Solution {
public:
vector<int> findRedundantConnection(vector<vector<int>>& edges) {
const int N = 1001;
m_ = vector<int>(N, 0);
for (int i = 0; i < N; ++i) {
m_[i] = i;
}
for (auto edge : edges) {
if (!u(edge[0], edge[1]))
return edge;
}
return {0, 0};
}
private:
vector<int> m_;
int f(int a) {
if (m_[a] != a)
m_[a] = f(m_[a]);
return m_[a];
}
bool u(int a, int b) {
int fa = f(a);
int fb = f(b);
if (fa == fb)
return false;
m_[fa] = b;
return true;
}
};
``````

## # 日期

2018 年 5 月 28 日 —— 太阳真的像日光灯～ 2019 年 1 月 25 日 —— 这学期最后一个工作日